Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $x = \dfrac{6z + 48}{z^2 + 5z - 24} \times \dfrac{z - 3}{-4z + 36} $
Solution: First factor the quadratic. $x = \dfrac{6z + 48}{(z - 3)(z + 8)} \times \dfrac{z - 3}{-4z + 36} $ Then factor out any other terms. $x = \dfrac{6(z + 8)}{(z - 3)(z + 8)} \times \dfrac{z - 3}{-4(z - 9)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ 6(z + 8) \times (z - 3) } { (z - 3)(z + 8) \times -4(z - 9) } $ $x = \dfrac{ 6(z + 8)(z - 3)}{ -4(z - 3)(z + 8)(z - 9)} $ Notice that $(z + 8)$ and $(z - 3)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ 6(z + 8)\cancel{(z - 3)}}{ -4\cancel{(z - 3)}(z + 8)(z - 9)} $ We are dividing by $z - 3$ , so $z - 3 \neq 0$ Therefore, $z \neq 3$ $x = \dfrac{ 6\cancel{(z + 8)}\cancel{(z - 3)}}{ -4\cancel{(z - 3)}\cancel{(z + 8)}(z - 9)} $ We are dividing by $z + 8$ , so $z + 8 \neq 0$ Therefore, $z \neq -8$ $x = \dfrac{6}{-4(z - 9)} $ $x = \dfrac{-3}{2(z - 9)} ; \space z \neq 3 ; \space z \neq -8 $